an inequality with noninteger order I want to Show that the following inequality is true or not:
For $0<q<1$ and $a\geq b\geq0$, $$a^q-b^q\leq 2(a-b)^q.$$
Could you please help me in showing this inequality is true or not? Because I have really no idea how can I approach.
Thanks in advance!...
 A: Hint:
Assuming $b \neq 0$ (because with $b=0$, the inequality is obviously true). We can re-write it as 
$$\left(\left(\frac{a}{b}\right)^q-1\right) \leq 2\left(\frac{a}{b}-1\right)^q.$$
Let $x=\frac{a}{b}$. Since $a \geq b$, therefore we only need to consider $x \geq 1$.
Consider the function $f(x)=2(x-1)^q-x^q+1$ on the interval $[1, \infty)$. Think about checking the monotonicity of this function. This will help you decide if the inequality holds.   
A: For  the case $a=b$ the inequality is trivial. Let $a\gt b$ and  $a-b=t$, then $a=t+b$ and put: $f(t)=2t^q-(t+b)^q+b^q \,\,\,\,$for $t\gt0$ we should show that $f(t)\gt 0$. 
We have :
$f'(t)=q(2t^{q-1}-(t+b)^{q-1})=\frac{2}{t^{1-q}}-\frac{1}{(t+b)^{1-q}}$
Hence $t\lt t+b$,$\,\,$ $f'(t)\gt 0$. So $f$ is increasing on $(0,\infty)$  and the minimum of $f$ hold in $t=0$. and it's mean:
$a^q-b^q\le 2(a-b)^q$
A: If one wishes to use only manipulation:
Suppose $\displaystyle \require\color \color\red{\frac{a}{b}<2}$, and $a>b>0$ since otherwise your inequality is trivial. Then your inequality follows from $$\frac{a^q-b^q}{(a-b)^q}=\left(\frac{a}{a-b}\right)^q-\left(\frac{b}{a-b}\right)^q=\left(1-\frac{b}{a}\right)^{-q}-\left(\frac{a}{b}-1\right)^{-q}\le\frac{a}{b},$$ which, setting $\displaystyle c=\frac{a}{b},$ becomes $$\left(1-\frac{1}{c}\right)^{-q}-(c-1)^{-q}=\frac{(c-1)^{-q}}{c^{-q}}-(c-1)^{-q} \le c.$$ Multiplying both sides by $(c-1)^q$ we get $$c^q-1\le c(c-1)^q. \tag{1}$$ Now, $c>1$ and $\require\color \color\red{c-1<1}$ respectively yield $c^q<c$ and $(c-1)^q>c-1$, so $(1)$ is implied by $$c-1<c(c-1) \\ c>1.$$
If instead $\displaystyle \require\color \color\green{\frac{a}{b}\ge2}$, let us start adding $2b^q$ to both sides of your inequality. This gives us $$a^q+b^q\le2(a-b)^q+2b^q \\ a^q+b^q\le 2( (a-b)^q + b^q ) $$ $$ \frac{a^q+b^q}{(a-b)^q+b^q}\le 2,$$ and this is weaker than $$\frac{a^q}{(a-b)^q}\le2 $$ $$ 1-\frac{b}{a}\ge 2^{-1/q}.\tag{2}$$ Finally, $(2)$ holds because $\displaystyle \frac{1}{2}$ is $\require\color \color\green{\text{between}}$ its LHS and RHS, and thus we're done.
